I've got some difficulties trying to understand the equation of state derived from Friedmann equations. I'd greatly appreciate it if someone walked me through this.(adsbygoogle = window.adsbygoogle || []).push({});

Now if the equation of state is stated as:

[tex]\Large \dot{\rho}+(3\rho +p)\frac{\dot{R}}{R}=0 \ \ |p=\omega \rho[/tex]

Then (in the case of pressure being zero):

[tex]\Rightarrow \rho \propto R^{-3} \Rightarrow \rho = \rho _0 (\frac{R_0}{R})^3[/tex]

I suspect the latter to be correct as it's not a result of my own logic

Now what I do not understand is the proportionality. If the equation of state is integrated I get something like this (set p=0):

[tex]\Large \dot{\rho}+(3\rho)\frac{\dot{R}}{R}=0

\Rightarrow \dot{\rho}=-3\rho\frac{\dot{R}}{R} \Rightarrow

\frac{1}{\rho}\dot{\rho}=-3\frac{\dot{R}}{R} \Rightarrow

\frac{1}{\rho}\frac{d\rho}{dt}=-3\frac{1}{R}\frac{dR}{dt}\ \|\cdot dt[/tex]

[tex]\Rightarrow

\int _{\rho _0}^\rho \frac{1}{\rho}d\rho}=-3\int _{R_0}^R \frac{1}{R}dR \Rightarrow

ln(\rho)-ln(\rho _0)=-3(ln(R)-ln(R_0)) \Rightarrow

ln \frac{\rho}{\rho _0}=-3ln\frac{R}{R_0}\Rightarrow

\frac{\rho}{\rho _0}=e^-3\frac{R}{R_0} \Rightarrow

\rho=e^-3\frac{R}{R_0}\rho _0[/tex]

Now is there some part to the theory that causes the equation to flip so that

[tex]\rho =e^-3\frac{R}{R_0}\rho _0 \Rightarrow \rho = \rho _0 (\frac{R_0}{R})^3[/tex]

or don't I just get the mathematics right?

Or have I done something wrong right in the beginning deriving the equation of state?

Edit:

Of course, how could I not see it... forgetting that a log x = log x^{a}.

Kurdt already told me that here, but then his message disappeared.

Well thanks to Kurdt anyway!

**Physics Forums - The Fusion of Science and Community**

# The equation of state

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: The equation of state

Loading...

**Physics Forums - The Fusion of Science and Community**